Question: Simplify and expand the following expression: $ \dfrac{3}{4x - 36}- \dfrac{5}{2x - 4}- \dfrac{2x}{x^2 - 11x + 18} $
Solution: First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor a $4$ out of denominator in the first term: $ \dfrac{3}{4x - 36} = \dfrac{3}{4(x - 9)}$ We can factor a $2$ out of denominator in the second term: $ \dfrac{5}{2x - 4} = \dfrac{5}{2(x - 2)}$ We can factor the quadratic in the third term: $ \dfrac{2x}{x^2 - 11x + 18} = \dfrac{2x}{(x - 9)(x - 2)}$ Now we have: $ \dfrac{3}{4(x - 9)}- \dfrac{5}{2(x - 2)}- \dfrac{2x}{(x - 9)(x - 2)} $ The least common multiple of the denominators is: $ 8(x - 9)(x - 2)$ In order to get the first term over $8(x - 9)(x - 2)$ , multiply by $\dfrac{2(x - 2)}{2(x - 2)}$ $ \dfrac{3}{4(x - 9)} \times \dfrac{2(x - 2)}{2(x - 2)} = \dfrac{6(x - 2)}{8(x - 9)(x - 2)} $ In order to get the second term over $8(x - 9)(x - 2)$ , multiply by $\dfrac{4(x - 9)}{4(x - 9)}$ $ \dfrac{5}{2(x - 2)} \times \dfrac{4(x - 9)}{4(x - 9)} = \dfrac{20(x - 9)}{8(x - 9)(x - 2)} $ In order to get the third term over $8(x - 9)(x - 2)$ , multiply by $\dfrac{8}{8}$ $ \dfrac{2x}{(x - 9)(x - 2)} \times \dfrac{8}{8} = \dfrac{16x}{8(x - 9)(x - 2)} $ Now we have: $ \dfrac{6(x - 2)}{8(x - 9)(x - 2)} - \dfrac{20(x - 9)}{8(x - 9)(x - 2)} - \dfrac{16x}{8(x - 9)(x - 2)} $ $ = \dfrac{ 6(x - 2) - 20(x - 9) - 16x} {8(x - 9)(x - 2)} $ Expand: $ = \dfrac{6x - 12 - 20x + 180 - 16x}{8x^2 - 88x + 144} $ $ = \dfrac{-30x + 168}{8x^2 - 88x + 144}$ Simplify: $ = \dfrac{-15x + 84}{4x^2 - 44x + 72}$